Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long inline gcd(long long x, long long y) {
  return !y ? (long long)abs(x) : gcd(y, x % y);
}
long long inline lcm(long long a, long long b) {
  return ((long long)abs((a / gcd(a, b)) * b));
}
double inline cartesian_dist(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long bigmod(long long b, long long p, long long m) {
  if (!p)
    return 1 % m;
  else if (p & 1)
    return (((b % m) + m) % m * bigmod(b, p - 1, m)) % m;
  else {
    long long ret = bigmod(b, p / 2, m);
    return (ret * ret) % m;
  }
}
long long inverse(long long a, long long m) { return bigmod(a, m - 2, m); }
long long extended(long long a, long long m) {
  long long oldr = a, r = m, oldans = 1, ans = 0, quotient;
  while (r) {
    quotient = oldr / r;
    oldr = oldr - r * quotient;
    swap(oldr, r);
    oldans = oldans - ans * quotient;
    swap(oldans, ans);
  }
  return ((oldans % m) + m) % m;
}
const double PI = acos(-1.0);
const double eps = 10E-12;
int n, m;
int given[100 + 2][100 + 2];
int cp[100 + 2][100 + 2];
int my[100 + 2][100 + 2];
int minCol[100 + 2];
int minRow[100 + 2];
vector<string> ans;
void rowClear() {
  for (int i = 1; i <= n; i++) {
    minRow[i] = 1000;
    for (int j = 1; j <= m; j++) minRow[i] = min(minRow[i], given[i][j]);
    for (int j = 1; j <= m; j++) {
      my[i][j] += minRow[i];
      given[i][j] -= minRow[i];
    }
    for (int j = 1; j <= minRow[i]; j++) {
      char s[10];
      sprintf(s, "row %d\n", i);
      ans.push_back(s);
    }
  }
}
void colClear() {
  for (int i = 1; i <= m; i++) {
    minCol[i] = 1000;
    for (int j = 1; j <= n; j++) minCol[i] = min(minCol[i], given[j][i]);
    for (int j = 1; j <= n; j++) {
      my[j][i] += minCol[i];
      given[j][i] -= minCol[i];
    }
    for (int j = 1; j <= minCol[i]; j++) {
      char s[10];
      sprintf(s, "col %d\n", i);
      ans.push_back(s);
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &given[i][j]);
      cp[i][j] = given[i][j];
    }
  }
  if (n <= m) {
    rowClear();
    colClear();
  } else {
    colClear();
    rowClear();
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (cp[i][j] != my[i][j]) {
        printf("%d", -1);
        return 0;
      }
    }
  }
  printf("%d", (int)ans.size());
  printf("\n");
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  return 0;
}